Math 224: Integral Calculus of One Variable FunctionsNorthwestern University, Lecture Notes

Written by Santiago Cañez

These are notes which provide a basic summary of each lecture for Math 224, “Integral Calculusof One Variable Functions”, taught by the author at Northwestern University. The book used asa reference is the 2nd edition of Essential Calculus: Early Transcendentals by Stewart. Watch outfor typos! Comments and suggestions are welcome.

Contents

Lecture 1: Areas and Distances 1

Lecture 2: The Definite Integral 5

Lecture 3: Evaluating Definite Integrals 10

Lecture 4: The Fundamental Theorem of Calculus 16

Lecture 5: Computing Integrals via Substitution 23

Lecture 6: Integration by Parts 29

Lecture 7: Trigonometric Integrals 34

Lecture 8: Trigonometric Substitution 39

Lecture 9: Partial Fractions 44

Lecture 10: Approximate Integration 48

Lecture 11: Areas Between Curves, Arclength 52

Lecture 12: Volumes 58

Lecture 13: Improper Integrals 61

Lecture 14: Sequences 66

Lecture 15: More on Sequences 70

Lecture 16: Series 74

Lecture 17: Integral Test 80

Lecture 18: Comparison Test 86

Lecture 19: Limit Comparisons and Alternating Series 89

Lecture 20: Ratio Test and Absolute Convergence 93

Lecture 21: Power Series 97

Lecture 22: Representing Functions as Power Series 103

Lecture 23: Taylor and Maclaurin Series 110

Lecture 24: More on Taylor Series 116

Lecture 25: Taylor Polynomials and Approximations 120

Lecture 1: Areas and Distances

Overview. In this course we will continue the study of single-variable calculus, focusing on thenotions of integrals and series. There are two overarching themes behind both of these topics: firstthe power of using approximations to solve concrete problems, and second the idea of making senseof what it means to add together infinitely many quantities. Indeed, integrals are defined via anapproximation procedure but in the end should be thought of as a type of “infinite summation”,while series are defined directly as a type of infinite summation but in the end can be used to givegood approximations to complicated functions. It’s not expected that you know what any of thismeans at this point, but is an idea we’ll highlight again and again.

Motivation. In the most basic formulation, integrals are used to compute areas. Of course, certainareas don’t require anything fancy to compute—say the area of a triangle or rectangle—but otherareas require more care.

For instance, you have likely seen that the area of a unit circle is π, which is a number whosedecimal expression looks like

π = 3.1415926536 . . . .

Of course, nowadays we can just have our calculator or compute tell us that this is what π is, buthow on earth could we figure this out without having such tools at our disposal? In particular, πwas known to the Ancient Greeks—how did they come to know that such a number existed? Thekey point is that we can actually approximate the area of a unit circle using areas of simpler-to-understand figures, and doing so is precisely what lead the Ancient Greeks to discover π.

To see how this works, imagine a unit circle surrounded by, say, a pentagon and also enclosinga pentagon:

The areas of these pentagons are possible, although somewhat tedious, to work out by hand, sayby dividing the pentagon into five triangles. Given the way in which these shapes are arranged wecan see that

area of small pentagon ≤ area of circle ≤ area of large pentagon.

But now, there is no reason to stop with pentagons! We can do the same thing using figures withmore sides:

2

The idea is that as the number of sides increases, the closer the figures come to approximating thecircle, and at each step of the way we have:

area of small n-sided figure ≤ area of circle ≤ area of large n-sided figure.

As n gets larger and larger, the term on the left gets closer and closer to the actual area of thecircle, as does the term on the right. In fact, after taking a limit as n goes to infinity, the leftmostterm approaches the same value as does the rightmost term, and this value is precisely π. Sincethe area of the circle is always sandwiched between these values, the area of the circle must bethis common value π. This limiting process where we approximate the area of the circle usingmultisided figures is indeed how the Ancient Greeks discovered π, and gives a glimpse of how wecan use approximations to derive concrete information.

Areas via rectangles. The type of area we will be interested in for now is the area of a regionlying under the graph of a function:

We will define integrals in more detail later on, but just to introduce the terminology now, thisarea is essentially what the so-called integral of f from a to b will give us. The key idea is that wecan determine this area through some limiting process where we approximate it using rectangles:

3

We will talk about this “limiting process” next time, and so will focus today on the approximationaspect, which is easiest to do by working through an example.

Example. Consider the function f(x) = 1− x3, whose graph looks like:

We want to eventually compute the area of the region under the graph of this function and betweenx = 0 and x = 1, which is the portion shaded in red above. Let us see how we can approximate thisarea using rectangles, which are simple geometric figures whose areas are possible to write downexplicitly.

Take the entire interval between x = 0 and x = 1 on the x-axis, and divide it into 3 subintervalsof equal length, so in this case the intervals from x = 0 to x = 1/3, x = 1/3 to x = 2/3, andx = 2/3 to x = 1. We use each of these subintervals as the base of a rectangle whose height is givenby the value of the function f at some point within that subinterval. For instance, we first use thevalues at the right endpoint of each subinterval as the height:

so the first rectangle has height given by the value of f(1/3), the second height given by f(2/3),and the third height given by f(1). Note that in this case it just so happens that f(1) = 0, so thefinal rectangle has no height and is thus represented by a “flat” line between x = 2/3 and x = 1.Nonetheless, we’ll still include this (not really a) rectangle in our equations to illustrate the generalprocess.

For each rectangle, its area is given by

f(right endpoint of subinterval) · (length of subinterval).

In this case we get

f

1

3

1

3=

1− 1

33

1

3f

2

3

1

3=

1− 2

3

33

1

3f(1)

1

3= (1− 1)1

3

4

as the areas of the three rectangles. The sum of these areas is meant to approximate the area ofthe region we want, and this sum is:

1− 1

33

1

3+

1− 2

3

33

1

3+ (1− 1)1

3, which is approximately 0.556.

Note that in this case, this value underestimates the area we want since the rectangles in questionare fully enclosed without our region but don’t cover all of it; this means that

0.556 ≤ area we want.

Now, there was nothing special about using right endpoints to give the heights of our rectangles,and we could just as well have used left endpoints instead. In this case we get rectangles whichlook like (in green):

with the previous rectangles still in red. In this case, the sum of the areas of the three greenrectangles is:

f(0)1

3+ f

1

3

1

3+ f

2

3

1

3= (1− 0)1

3+

1− 1

33

1

3+

1− 2

3

33

1

3,

which is approximately 0.889. In this case, this value overestimates the area in question since thegreen rectangles always go above the region in question; this gives

area we want ≤ 0.889,

which together with the previous inequality gives

0.556 ≤ area we want ≤ 0.889.

Warning. Because of the decreasing nature of the function f(x) = 1 − x3, it turns out that inthis case using left endpoints to give heights of rectangles will always lead to an overestimationof the area and using right endpoints will always lead to an underestimation. However, this doesnot happens for all functions; for instance, with an increasing function the opposite would be true:left endpoints would underestimate and right endpoints will overestimate. For functions which areneither increasing nor decreasing, other things can happen.

Taking a limit. Going back to our example, the idea is that as we carry out the same procedureonly using more and more rectangles, obtained by splitting our original interval from x = 0 to

5

x = 1 into more and more pieces, the values we get will provide better and better approximationsto the area we want. For instance, if we had used n = 4 subintervals:

x = 0 to x =1

4, x =

1

4to x =

2

4, x =

2

4to x =

3

4, x =

3

4to x = 1,

using left endpoints to give heights leads to rectangles whose areas add up to:

f(0)1

4+ f

1

4

1

4+ f

2

4

1

4+ f

3

4

1

4= (1− 0)1

4+

1− 1

43

1

4+

1− 2

3

43

1

4+

1− 3

4

44

1

4,

which is approximately 0.859. The actual area of the region in question is 0.75 (we’ll see how todefinitely compute this later), so the 0.859 value we got using left endpoints and 4 subintervals is abetter approximation than the 0.889 value we got using only 3 subintervals. Using right endpointsand 4 subintervals gives an approximate value of 0.609, which is closer to the actual 0.75 than the0.556 from before. To get the actual area we would continue this process, taking n subintervalswith n getting larger and larger, eventually taking a limit as n goes to infinity.

This process works (!), but the problem is that limit expression obtained in this final step is verydifficult to work with in practice, and it is only in certain circumstances that it becomes directlycomputable. Nonetheless, this limiting process is what we’ll use to define integrals soon enough,but the upshot is that this is NOT the method we’ll use to actually compute integrals. Magically,it turns out that there is an alternate much more direct way of computing values of integrals usingantiderivatives, and it is this discovery which made Isaac Newton famous in the 17th century. Thisis the content of the Fundamental Theorem of Calculus, which we’ll talk about next week. For now,we are simply highlighting how we can use rectangles to approximate areas, foreshadowing otherapproximation techniques we’ll see later on.

Distance. We finish with what seems to be totally unrelated question to the one we were consid-ering above, but turns out to be a reflection of the same idea. Suppose Usain Bolt (I’ve still gotthe Olympics on my mind!) is running and someone measures his speed (say in meters per second)every 5 seconds for 20 seconds:

time 5 10 15 20

speed 10 15 13 8

The problem is to estimate how much distance Usain traveled over these 20 seconds. Knowing that

distance = speed times time,

we can estimate that over the first 5 seconds he traveled

10 · 5 = 50 meters,

after the next 5 seconds he traveled

15 · 5 = 75 meters,

and so on. Overall he covered a distance of approximately

10 · 5 + 15 · 5 + 13 · 5 + 8 · 5 = 230 meters.

(So, this seems unrealstic: Usain Bolt’s world record time for 200 meters is just under 20 seconds,and it seems unlikely he could travel the remaining 30 meters in under 1 second as our numberswould suggest, but oh well.)

6

The point to recognize is that the type of sum we ended up with above:

10 · 5 + 15 · 5 + 13 · 5 + 8 · 5

is similar to the type of sum we saw previously when approximating areas using rectangles. Indeed,if we set up a pair of axes where the horizontal axis is time and the vertical axis speed:

the sum above can be interpreted as the sum of the areas of the rectangles shown here. The upshotis that this problem, although not directly phrased in terms of areas, can in fact be interpreted insuch a way, and suggests that integration in general has many applications which go way beyondarea problems. Indeed, we’ll see that distance can indeed be obtained by integrating speed, andthe sum we used above to approximate such a distance is at the core of the reason why.

Lecture 2: The Definite Integral

Warm-Up. We estimate the area under the graph of the function drawn below between x = 1and x = 5:

We do so using rectangles obtained by dividing the interval from x = 1 to x = 5 into four pieces ofequal length, and using midpoints to determine the heights. Thus we use the following rectangles:

7

The sum of the areas of these rectangles is

f(1.5)(1) + f(2.5)(1) + f(3.5)(1) + f(4.5)(1),

where as stated we use the value of f at the midpoint of each subinterval as the height. Based onthe given graph, these values are:

f(1.5) = 5 f(2.5) = 8 f(3.5) = 7 f(4.5) = 10,

so the sum of the areas of the rectangles is:

5(1) + 8(1) + 7(1) + 10(1) = 30.

Thus we approximate the area under the graph of the given function between x = 1 and x = 5 asbeing close to 30. Using more rectangles should give better estimates.

A modification. Here’s a seemingly different question. Suppose water is being added to a pool,and that the rate at which the water flows in at a given time is given by the function:

Now we interpret the x axis as time (say in seconds) and the vertical axis as a rate. So, the valuef(x) gives the rate at which water flows in at time x. We want to estimate the total amount ofwater which went into the pool between x = 1 and x = 5 seconds.

The point is that if water flows in at a constant rate over a period of some number of seconds,the amount which flowed in over that time period is

amount = rate · time.

If we use the time intervals x = 1 second to x = 2 seconds, x = 2 seconds to x = 3 seconds, and soon, and we use the half-way points in each such subinterval as the estimate for the rate over that

8

corresponding subinterval, then we estimate the total amount of water which flows in between 1and 5 seconds to be:

f(1.5)(1) + f(2.5)(1) + f(3.5)(1) + f(4.5)(1) = 5(1) + 8(1) + 7(1) + 10(1) = 30.

The observation of course is that the sum on the left is the same sum we used in the Warm-Up areaproblem, the connection being that in both cases we are adding up various products. As mentionedlast time, this suggests that integrals have wide applications going way beyond computing areasalone.

An explicit area. Let us now try to compute a precise area, no longer approximate areas. Indeed,let’s return to the function f(x) = 1− x3 we used last time:

and actually work out the concrete area of the portion in red between x = 0 and x = 1. The ideais one we’ve mentioned: we approximate this using rectangles, but then take a limit as the numberof rectangles grows larger and larger. For some number n, we divide the interval from x = 0 tox = 1 into n subintervals of equal length. Since our entire interval has length 1, this means thateach subinterval will have length 1n , and so the subintervals we use are:

x = 0 to x =1

n, x =

1

nto x =

2

n, x =

2

nto x =

3

n, and so on,

with the final subinterval being the one from x = n−1n to x =nn = 1. We will use left endpoints

when determining the heights of rectangles, so the rectangles we use look like:

The left endpoints of these given subintervals are 0, 1n ,2n , . . . ,

n−1n , so the sum of the areas of the

rectangles is

f(0)1

n+ f

1

n

1

n+ f

2

n

1

n+ · · ·+ f

n− 1n

1

n,

9

where the · · · indicate a bunch of intermediate terms.

Riemann sums. The type of expression we ended up with above:

f(0)1

n+ f

1

n

1

n+ f

2

n

1

n+ · · ·+ f

n− 1n

1

n

is known as a Riemann sum for the function f . Geometrically, a Riemann sum is nothing but thesum of areas of a bunch of triangles. Now, Riemann sums come in all shapes and flavors, dependingon which specific subintervals are used and which “sample points” we use to give the heights of therectangles in question. For instance, in general there is no reason why we couldn’t use subintervalsof unequal length when estimating areas, and there is no reason why we couldn’t use a left endpointfor one subinterval, a midpoint for another, and some other random point from another. Any suchsum obtained by such choices is still a Riemann sums, it’s just that certain Riemann sums lead tonicer expression.

You can check the book for more formal notation and terminology (such as the notion of apartition) associated with Riemann sums, but we won’t dwell too much on such things. After all,we’ll soon get to a point where we can compute these areas without considering Riemann sums atall, so for now we’re just getting a glimpse as to how Riemann sums work.

Definite integrals. The idea, then, is to consider Riemann sums as the width of the rectanglesused becomes smaller and smaller, which in turn will produce more and more rectangles. As thewidths approach zero (or correspondingly the number of rectangles goes off to infinity), the valuesobtained give better and better approximations to the area under the graph of the function, so thatthis area is obtained as a limit of such Riemann sums.

In general, given a function f and an interval from x = a to x = b, we define the definite integralof f from x = a to x = b to be the limit of Riemann sums for f over the interval x = a to x = b asthe widths of the subintervals used goes to zero:

b

af(x) dx = limit of Riemann sums as widths go to zero.

The notation on the left is our notation for such a definite integral, where the lower and upper“bounds” on the integral symbol (the elongated S symbol at the beginning) indicate the intervalx = a to x = b we’re looking at. For now the “dx” should be viewed as simply part of the notation,but we’ll see what it is meant to stand for soon. The book has this definition written out in moreformal notation, but the core idea is what I wrote above. Geometrically, such integrals preciselycompute the areas we’ve been considering:

b

af(x) dx = “area” between the graph of f and the x-axis between x = a and x = b.

We’ll see later what I mean by “area”. (The point is that the value can actually be negative.)

Back to explicit area. Returning to our example, we’re looking at the Riemann sum

f(0)1

n+ f

1

n

1

n+ f

2

n

1

n+ · · ·+ f

n− 1n

1

n.

We have a nice way of expressing such sums using more compact notation, so called “sigma”notation:

n−1

k=0

f

k

n

1

n.

10

The

symbol is the Greek letter “sigma”, and should be thought of as an “S” for “Sum”. Thisnotation indicates the sum of the terms

f

k

n

1

n

as k varies from k = 0 to k = n− 1, increasing the value of k by 1 at each step. So, the first termwhen k = 0 would be

f(0)1

n,

the second term when k = 1 is

f

1

n

1

n,

the term when k = 2 is

f

2

n

1

n,

and so on, with the final term when k = n− 1 being

f

n− 1n

1

n.

The sum of all these is indeed the Riemann sum we were considering before:

n−1

k=0

f

k

n

1

n= f(0)

1

n+ f

1

n

1

n+ f

2

n

1

n+ · · ·+ f

n− 1n

1

n.

Sigma notation is often simpler to work with than the expression on the right with the ambiguous-looking · · · occurring in the middle.

Now, the actual function we’re looking at is f(x) = 1− x3, so

f

k

n

= 1− k

3

n3.

Thus our Riemann sum is actually:n−1

k=0

1− k

3

n3

1

n.

As n goes off to infinity, the widths of our subintervals goes to zero, and so the definition of thedefinite integral we gave above would say that:

1

0(1− x3) dx = lim

n→∞

n−1

k=0

1− k

3

n3

1

n.

Our goal is now to figure what this sum actually is, so that we can then actually determine thislimit, and hence the area we want, precisely.

We can write our Riemann sum as

n−1

k=0

1− k

3

n3

1

n=

n−1

k=0

1

n− k

3

n4

11

after we distribute the 1n term to both terms in the parenthesis. Recall that this sigma notationdenotes the sum:

1

n− 0

3

n4

+

1

n− 1

3

n4

+

1

n− 2

3

n4

+ · · ·+

1

n− (n− 1)

3

n4

.

Now, all together in this sum we have n copies of 1n , one for each of the subintervals we had.Grouping these terms together gives n 1n = 1, and grouping the remaining terms together gives

− 03

n4− 1

3

n4− 2

3

n4− · · ·− (n− 1)

3

n4= − 1

n4(03 + 13 + 23 + · · ·+ (n− 1)3).

Thus our Riemann sum simplifies to:

n−1

k=0

1− k

3

n3

1

n=

n−1

k=0

1

n− k

3

n4

= 1− 1

n4(03 + 13 + 23 + · · ·+ (n− 1)3)).

Now, it is a fact that

03 + 13 + 23 + · · ·+ (n− 1)3 =n−1

k=0

k3 =(n− 1)2n2

4.

This is not at all obvious and requires a good amount of work to justify; this is the type of thingwhich you would have to be given as a fact if you were to be required to use it since no one in theirright mind (except for mathematicians) would know this off the top of their head. Ask in officehours if you are really interested in seeing where this comes from. The upshot is that we can finallywrite our Riemann sum as:

1− 1n4

(03 + 13 + 23 + · · ·+ (n− 1)3)) = 1− 1n4

(n− 1)2n24

.

After rewriting this expression so that everything uses the same denominator and simplifying, weget:

1− 1n4

(03 + 13 + 23 + · · ·+ (n− 1)3)) = 4n4 − (n− 1)2n2

4n4=

3n4 + 2n3 − n24n4

,

so that finally after all this work we get that:

1

0(1− x3) dx = lim

n→∞

3n4 + 2n3 − n24n4

.

WHEW!!! That was a lot of work! But fear not, the point is that this should seem tedious andshould make you hope that there is a simpler way of computing integrals, which there is (!), aswe’ll talk about next week. But, going through this type of thing at least one is good for getting asense of where things actually come from, and for making you appreciate just how important theFundamental Theorem of Calculus we’ll look at next week actually is.

The final limit obtained is now one we can compute using ideas from Math 220. For instance,by dividing the numerator and denominator both by n4 we can rewrite this limit as

limn→∞

3n4 + 2n3 − n24n4

= limn→∞

3 + 2n −1n2

4,

12

which is thus equal to 34 since the2n and

1n2

terms both go to zero as n goes to infinity. Thus, wefinally get our desired value, the area of the region we’ve been considering is

1

0(1− x3) dx = 3

4.

We mentioned this value (or rather, 0.75) in passing last time, when assessing how good theestimates we found for this area actually were. Now we know where it came from.

Integrals via geometry. The fact that integrals can be interpreted as areas makes certainintegrals easily computable now, using what we already know about geometry. For instance, theintegral 1

0x dx

gives the area of the following triangular region, below the line y = x between x = 0 and x = 1:

This triangle has area 12 , so 1

0x dx =

1

2.

The integral 2

1x dx

gives the area of the region:

13

By thinking of this as a square of height 1 with a triangle on top we can see that this region hasarea 32 , so 2

1x dx =

3

2.

Finally, consider the integral 1

−1

1− x2 dx.

The curve y =√1− x2 describes the top half of the unit circle, which we can see by squaring both

sides to get y2 = 1 − x2 and then rearranging to get x2 + y2 = 1. We get the to half since weare taking the positive square root of 1− x2; the bottom half would have equation y = −

√1− x2.

Thus the integral above gives the area of the region:

This is half the region enclosed by the entire unit circle, so since the unit circle has area π, thisparticular region has area π2 , so 1

−1

1− x2 dx = π

2.

Of course, not all integrals are so easily computable, since the areas in question are not always onesof well-known geometric figures. We’ll see next week the “real” way to compute such integrals.

Lecture 3: Evaluating Definite Integrals

Warm-Up. We determine the value of the limit

limn→∞

n

k=1

1

n

1− k

2

n2

by interpreting it as an area via an integral. (Note that the original version I gave in class used 2ninstead of 1n and

2kn instead of

kn . I’ll remind you below why I changed this to the version given

here.) The point is that this limit is pretty much impossible to compute directly, but by recognizingwhat it is actually mean to represent, it becomes much more manageable. The types of limits ofRiemann sums which define integrals look like

limn→∞

n

k=1

f(some value)(width).

So, in order to interpret the given limit as such an expression, we must determine what is the“width” term, what is the function being considered, and what is the interval being integrated over.

14

The 1n term in our case plays the role of “width”. Thekn terms represent sample points from the

subintervals being considered. In this case, as k increases from k = 1 to k = n, these kn same pointsrun through the values

1

n,2

n,3

n, . . . ,

n

n= 1.

Thus we are looking at the interval [0, 1] being divided into n pieces of equal length, using rightendpoints as sample points. (Using left endpoints would have corresponded to 0, 1n ,

2n , . . . ,

n−1n ,

which would have arisen if the sum in question ran from k = 0 to k = n− 1.)Since the 1n in the original summation thus corresponds to the lengths of the subintervals ob-

tained by breaking [0, 1] into n pieces of equal length, the remaining

1− k2n2

term must correspond

to thef(some value)

piece which gives the height of a rectangle making up our Riemann sum. Thus, the function beingconsidered is

f(x) =

1− x2,

and note that evaluating this at the given right endpoints kn indeed gives

f

k

n

=

1− k

2

n2

as we need. (The reason why using 2n and2kn as I originally had is bad is that this would correspond

to the interval [0, 2], since when k = n the final right endpoints 2kn would be 2. However, the function

f(x) =√1− x2 is not defined for x > 1 since these values give a negative term under the square

root. This is why I modified the original sum so that only the interval [0, 1] would be considered.)The overall point is that, for the function f(x) =

√1− x2, the expression

1

n

1− k

2

n2corresponds to (width)(f evaluated at right endpoints),

so that the sumn

k=1

1

n

1− k

2

n2

is a Riemann sum for f(x) =√1− x2 over the interval [0, 1]. Thus, the limit in question is precisely

the integral of this function over the interval [0, 1]:

limn→∞

n

k=1

1

n

1− k

2

n2=

1

0

1− x2 dx.

The upshot is that this is an integral we can compute geometrically, based on its interpretationas an area. Indeed, we saw last time that the graph of f(x) =

√1− x2 is the upper-half of the unit

circle, so here we are considering the area under this upper-circle from x = 0 to x = 1, which isthus a quarter of a circle overall:

15

(The area of the rectangle in green is given by one of the terms 1n

1− k2

n2in our Riemann sum.)

The full unit circle has area π, so our region has area π4 . Hence

limn→∞

n

k=1

1

n

1− k

2

n2=

1

0

1− x2 dx = π

4

is the desired value of the limit on the left.

Net area. The value of the integral 1

−1x dx

is zero, which seems to go against the idea that integrals compute areas. The point of clarification wenow make is that it is not literally true that integrals compute areas, but rather what they actuallycompute is “net area”. The key observation is that “areas” under the x-axis are actually countedas being negative, so an integral can in general have positive and negative area contributions, andwhat we get is a sum of these contributions. Indeed, the fact that “areas” below the x-axis arecounted as being negative comes directly from the Riemann sum approach: in an “area” term givenby

f(sample point)(width of subinterval),

the “height” f(sample point) can be negative depending on the function f , so that the productabove is not the actual area of some rectangle but rather the negative of this area.

In the example above, the region between the graph of f(x) = x and the x-axis between x = −1and x = 1 looks like:

16

This region consists of two triangles of equal area, only that one is counted as positive and one asnegative, so that adding them together gives zero:

1

−1x dx = 0.

Another way to phrase this is as follows. Splitting the interval entire interval [−1, 1] into two pieces[−1, 0] and [0, 1] gives a corresponding splitting of intervals:

1

−1x dx =

0

−1x dx+

1

0x dx.

Each integral on the right measures the net area of one of the triangles in the picture we’re lookingat, only that the first integral on the right gives negative the area (which is −12) and the secondgives positive the area (which is 12). Thus, the first integral is precisely the negative of the second,so their sum is zero: 1

−1x dx =

0

−1x dx+

1

0x dx = −1

2+

1

2= 0.

To summarize, we now have the true geometric meaning behind an integral: ba f(x) dxmeasures

the net area of the region between the x-axis and the graph of f between x = a and x = b, whereareas below the x-axis count as negative and areas above the x-axis count as positive.

Area function. Suppose f is the function whose graph is drawn below in red:

Define a new function F by setting the value of F at x to be

F (x) =

x

0f(t) dt.

In other words, based on our geometric interpretation of this integral, F (x) gives the net areabetween the graph of f and the x-axis from 0 to whatever value x is, where x is allowed to vary.Thus, F measures net area up to some given variable point. For instance, for the given green pointx, the area of the region shaded in green is F (x).

A word about the notation. Note that in the definition of F the integral in question uses t asthe variable, so that we have f(t) and dt. We should not use x here since we are already using xto denote the variable upper bound on the integral. The point is that x and t denote two separatethings: x the point which tells how far to integrate to, and t the variable of integration itself. Note

17

that the fact we use t to denote the variable of integration is irrelevant, and we could just haveeasily used any other letter or symbol instead. In other words,

b

af(t) dt,

b

af(u) du, and

b

af() d

all mean the same thing, as long as the variable used in the function f and the variable used in the“d” part at the end match up.

Now we ask some questions about the function F . For instance, is F (5) > 0? Is F (12) > 0?Are there any values of x where F (x) = 0? First, F (5) gives the net area between the graph off and the x-axis up to 5, which has a positive contribution coming from the first “hump” andthen a negative contribution. The area which gives the negative contribution seems to be largerthan the area which gives the positive contribution, so F (5) will be negative. Next, F (12) givesthe net area all the way to 12, which has two positive contributions and one negative contribution.Overall, from eyeballing the picture, it seems that the positive contribution areas are larger thanthe negative area contribution, so F (12) is positive.

Finally, note that of course F (0) = 0, since F (0) measures the area from 0 to 0, and there isno such area there. (Or, you can think of this as saying that the area of a single “point” is zero.)However, there is another value a bit before 5 indicated by the blue a at which F (a) will also bezero. Indeed, the net area up to this point has positive and negative contributions, and I’ve drawn(or at leat tried to) the point a at the point where these two effects would cancel each other out,meaning that area of the first hump above the horizontal axis should be the same as the area ofthe portion below the horizontal axis from the point where the graph intercepts that axis up to a.There will actually be another point (not drawn) a bit after the graph crosses the horizontal axisand goes positive again at which F (x) will again be zero. We’ll come back and see what is specialabout this type of “area function” next time.

Evaluation Theorem. We now come to what the book refers to as the Evaluation Theorem, andwhich most other sources refer to as the Fundamental Theorem of Calculus. (Our book eventuallyuses “Fundamental Theorem of Calculus, Part II” to refer to the Evaluation Theorem, to contrastit with a very much related fact which also goes by the name “Fundamental Theorem of Calculus”.We’ll get to this next time.) This is the fact which tells us that integrals can be computed in amuch simpler way than having to resort to the limit of Riemann sums definition, which is verydifficult to work with in general. The surprising (or “fundamental”) fact is that evaluating anintegral amounts to finding an antiderivative.

The statement is that if F is an antiderivative of f , meaning F is a function whose derivativeis f (i.e. F ′(x) = f(x) for all x), then

b

af(x) dx = F (b)− F (a).

Thus, in order to compute an integral, we first find an antiderivative of the function being integrated,and then we evaluate this antiderivative at our bounds and subtract. We’ll see antiderivatives arenot always straightforward to find, but for the most part they are simpler to find than the alternateapproach of trying to compute some limit of Riemann sums explicitly. The rest of the “integration”half of the course will essentially be devoted to techniques used to find antiderivatives. Notation-wise, the difference F (b)− F (a) showing up above is normally shortened to

F (b)− F (a) = F (x)b

a

or F (x)

baas the book uses

,

18

so that the Evaluation Theorem can be written as

b

af(x) dx = F (x)

b

a, where F ′(x) = f(x).

I referred above to the fact that the Evaluation Theorem is in many ways quite surprising, sincethe definition of an integral via rectangles seems to have nothing to do with derivatives, let aloneantiderivatives. The fact that the process of integration is so directly related to the process ofdifferentiation was a major discovery in the history of mathematics and is part of what made IsaacNewton famous in the 1600’s. People had been working with integral-like objects since the time ofthe Ancient Greeks, but before Newton (and a mathematician named Leibniz) came along, thereonly existed a hodge-podge of various techniques for actually evaluating such things, which onlyworked in certain situations. Newton and Leibniz’s realization (and reason why the “FundamentalTheorem of Calculus” is referred to as “fundamental”) was that all of these techniques could beencoded more efficiently via anti-differentiation, which was truly a magnificent accomplishment.If nothing else, I hope you come to appreciate how much “simpler” it is to compute integrals inthis manner than it is via the Riemann sum definition. (“Simpler” here is used in a relative sense;antiderivatives can still be tricky to compute in general.)

Example 1. We return to some examples we saw previously. Consider

1

0x dx.

According to the Evaluation Theorem, to compute this integral all we need to do is come up with afunction whose derivative is x. Thinking back to what we know about derivatives, we know that inorder to get x after differentiation we should start with something like x2. However, the derivativeof x2 is 2x, whereas we just want to end up with x. To deal with the extra factor of 2 at the frontof 2x, we should multiply our candidate antiderivative of x2 by 12 to get

1

2x2.

The derivative of this function is indeed just x, and so the Evaluation Theorem says that:

1

0x dx =

1

2x2

1

0.

Again, the notation on the right means “plug 1 into the function 12x2, plug 0 into it, and subtract”,

so that we get 1

0x dx =

1

2x2

1

0=

1

2(1)2 − 1

2(0)2 =

1

2

as the value of the integral in question. Note that this agrees with the value we found last time viageometry after interpreting this integral as the area of a triangle.

The same antiderivative also works when evaluating 21 x dx, which we also worked out last time

using geometry. The Evaluation Theorem gives

2

1x dx =

1

2x2

2

1=

1

2(2)2 − 1

2(1)2 =

3

2,

agreeing with the value we saw last time.

19

Example 2. We evaluate π

0sinx dx.

Again, all we need is a function whose derivative is sinx. We know that the derivative of cosxgives − sinx, so this almost works. To get what we want we simply multiply by an extra negative,so we see that − cosx is a function whose derivative is sinx. The Evaluation Theorem gives

π

0sinx dx = − cosx

π

0= (− cosπ)− (− cos 0) = 1− (−1) = 2.

(Be careful keeping track of the negatives!) Geometrically, this gives the area of the following regionunder the graph of f(x) = sinx:

which is an area which is pretty much impossible to determine without the use of an integral andthe Evaluation Theorem.

If instead we had π

05 sinx dx,

we would need a function whose derivative is 5 sinx. Above we saw that − cosx has derivativesinx, so −5 cosx will have derivative 5 sinx. This illustrates a general fact, namely that

b

acf(x) dx = c

b

af(x) dx

when c is a constant. That is, constants can be “pulled out” of integrals, similarly to how constantscan be pulled out of derivatives. Thus in this case we get:

π

05 sinx dx = 5

π

0sinx dx = −5 cosx

π

0= (−5 cosπ)− (−5 cos 0) = 10.

Example 3. Finally we determine the value of

π

0(5 sinx+ x2) dx.

So, we need an antiderivative of 5 sinx+ x2. The key point is that we already know that −5 cosxis an antiderivative of 5 sinx, so if we also had an antiderivative of x2, adding it to −5 cosx wouldgive an antiderivative of 5 sinx+ x2. This comes from the fact that

derivative of (f + g) = (derivative of f) + (derivative of g).

20

In terms of integrals, this corresponds to the fact that

b

a(f(x) + g(x)) dx =

b

af(x) dx+

b

ag(x) dx,

meaning that integrals of sums can always be split up into pieces. In our case, this means that π

0(5 sinx+ x2) dx =

π

05 sinx dx+

π

0x2 dx.

The first integral was computed to be 10 in Example 2. For the second we need a function whosederivative is x2; x3 almost works except that this gives us a 3 in front, so we throw in a 13 in frontto balance this out. Thus:

π

0(5 sinx+ x2) dx =

π

05 sinx dx+

π

0x2 dx

= −5 cosxπ

0+

1

3x3

π

0

= 10 +

1

3π3 − 1

303

= 10 +1

3π3.

Indefinite integrals. Since computing integrals comes down to finding antiderivatives, it wouldbe enormously helpful to have the basic antiderivatives engrained in our minds. These can be foundin Table 1 of Section 5.3, and are all ones which you eventually want to have memorized. We’lllook at more examples next time.

As a matter of notation, because of the connection between integrals and antiderivatives, weuse

f(x) dx

to refer to the possible antiderivatives of the function f . This is what is known as an indefiniteintegral, to distinguish it from the definite integrals we’ve seen up to now. The difference is that adefinite integral b

af(x) dx

gives a number as a result, while an indefinite integral

f(x) dx (with no bounds on the integral sign)

gives a collection of functions as a result, namely the functions which have f as their derivatives.You’ll see in the book that all of these basic indefinite integral formulas include a “+C” at the end,which comes from the fact that adding a constant to a function does not alter its derivative. Lateron we’ll see situations where this “+C” plays an important role.

Lecture 4: The Fundamental Theorem of Calculus

Warm-Up 1. We evaluate the integral

3

1(2ex − 2x6 + sec2 x) dx.

21

To do so we need to find an antiderivative of the integrand 2ex− 2x6+sec2 x. (The term integrandjust refers to the function which is being integrated.) First, 2ex is its own derivative:

(2ex)′ = 2ex.

Now, to get the x6 in the second term we need to take the derivative of x7:

(x7)′ = 7x6.

To get rid of the 7 in front we divide by 7 throughout:

1

7x7

′= x6,

and finally to get −2x6 as required we multiply through by −2:−27x7

= −2x6.

Finally, the derivative of tanx is sec2 x:

(tanx)′ = sec2 x.

Thus 2ex − 27x7 + tanx is an antiderivative of 2ex − 2x6 + sec2 x. Since adding any constant to

2ex − 27x7 + tanx still gives the same derivative (since the derivative of a constant is zero), we see

that the most general antiderivative of 2ex − 2x6 + sec2 x is

(2ex − 2x6 + sec2 x) dx = 2ex − 27x7 + tanx+ C

where C denotes an arbitrary constant. Recall that this integral notation without any boundsrefers to the indefinite integral of 2ex − 2x6 + sec2 x, which, by definition, means the most generalantiderivative of 2ex− 2x6+sec2 x. (Indefinite integrals always include this +C term, and later wewill see instances where this +C actually matters.)

To evaluate the original integral, all we need is an antiderivative of 2ex − 2x6 + sec2 x. Thuswe can just take the constant C to be zero; in other words, 2ex − 27x

7 + tanx is one possibleantiderivative of 2ex − 2x6 + sec2 x, which is all we need in order to evaluate a definite integral.Thus:

3

1(2ex − 2x6 + sec2 x) dx =

2ex − 2

7x7 + tanx

3

1

=

2e3 − 2

7· 37 + tan 3

−

2e− 2

7+ tan 1

.

Note that if had used, say, 2ex− 27x7+tanx+3 as an antiderivative of 2ex−2x6+sec2 x instead of

the one we used, the extra +3 term would have disappeared anyway since it would show up oncewhen we plug in the upper bound of 3 but then again when we plug in the lower bound of 1, sothat after subtracting the +3 would be no more. This is why the +C term in an indefinite integralwon’t matter when computing a definite integral.

Warm-Up 2. Suppose that water is flowing in/out of a tank at a rate of f(t) = 3t3 − 1t2

cubicmeters per second. We want to determine the total amount of water which flowed in/out of the

22

tank over the time period of t = 1 second to t = 4 seconds, or in other words the net change in thevolume over this time period. The point is that this is simply asking to compute the integral

4

1

3t3 − 1

t2

dt.

Indeed, we’ve alluded to the idea before that “net change” questions can also be phrased in termsof integrals (such as the type of distance question we saw in the first lecture), so that integrals arenot only meant to have applications to computing area. (Note that the version I first gave in classhad 0 as the lower bound of the integral, which is nonsense since the 1

t2portion of the function we

are integrating isn’t even defined at 0. This is why I changed the lower bound to 1 instead.)We can now see why this is true from the Evaluation Theorem. Write the Evaluation Theorem

as b

aF ′(x) dx = F (b)− F (a).

The right side is the net change in F from a to b. The F ′(x) on the left measures the rate ofchange of F , so this equation says that integrating the rate of change precisely results in the netchange. In our specific problem, f(t) = 3t3 − 1

t2is the rate of change of the volume function F (t),

so integrating it from 1 to 4 should give the net change in F itself.Note also something we mentioned last time: the fact that we are using t as the variable makes

no difference to the procedure for evaluating this integral, and the value of this integral is the sameas the one for 4

1

3x3 − 1

x2

dx, or

4

1

33 − 12

d.

We use t in this case simply because we are interpreting this variable as “time”.So, we compute the given integral. We have:

(t4)′ = 4t3, so

3

4t4′

= 3t3.

Now, to find an antiderivative for 1t2, we write this function as t−2 instead. Then the same differ-

entiation rules apply: to get a power of −2 we need to start with a power of −1:

(t−1)′ = −t−2.

Thus, 1t is an antiderivative for −1t2, so all together

3

4t4 +

1

tis an antiderivative for 3t3 − 1

t2.

The most general antiderivative, if we had been asked for it, is

3t3 − 1t2

dt =

3

4t4 +

1

t+ C.

To finish computing our specific integral, we have:

4

1

3t3 − 1

t2

dt =

3

4t4 +

1

t

4

1

=

3

4· 44 + 1

4

−3

4+ 1

23

=770

4.

Thus the net change in volume of water in the tank between 1 and 4 seconds was 7704 cubic meters.

Fundamental Theorem of Calculus. Recall the idea of an area function from last time: for afunction f , we can define a new function F by setting the value of F at x to be:

F (x) =

x

af(t) dt.

This value gives the net area between the graph of f and the horizontal axis from a fixed point a upto some variable point x. As x changes, this net area changes as well. The Fundamental Theoremof Calculus (or Part I of this theorem, to distinguish it from Part II, which is also known as theEvaluation Theorem), tells us what happens when we take the derivative of this area function. Theresult is that

for F (x) =

x

af(t) dt, we have F ′(x) = f(x).

Thus, differentiating the function obtained by integrating another function gives the function whichwas originally being integrated.

Now, at first glance this doesn’t seem to be saying much, since we’ve already said before inthe Evaluation Theorem that integrating a function should give its antiderivative, and of courseit’s true that differentiating this antiderivative gives the original function. The point, which we’llelaborate on next time, is that the reason why the Evaluation Theorem works and the reason whyintegrating a function amounts to finding an antiderivative is because of the Fundamental Theoremof Calculus. This specific version is called Part I since it logically is supposed to come before PartII. Again, we’ll elaborate on this next time.

Integration and differentiation do opposite things. Before looking at some examples of howto use the Fundamental Theorem of Calculus, we emphasize what it and the Evaluation Theoremreally say. The Evaluation Theorem says

b

aF ′(x) dx = F (b)− F (a)

and the Fundamental Theorem says

d

dx

x

af(t) dt

= f(x).

You should read the first as saying the following: start with a function F , take the derivative, andthen take the integral, and you get back the original F itself. The second equation above says: startwith a function f , take the integral, and then take the derivative, and you get back the originalf itself. The point is that these together say that integration and differentiation are “inverse”operations, meaning that one does the opposite of what the other one does.

Example 1. Define a function F by

F (x) =

x

0t3 dt.

24

The Fundamental Theorem of Calculus says that F ′(x) = x3. Note that in this case f is thefunction f(t) = t3, and the final conclusion that F ′(x) = f(x), where f(x) means that we evaluatef , not at t, but at the upper bound x of the integral in question. Saying that

F ′(x) = t3

is incorrect; we must always evaluate the result at the upper bound of the integral since this iswhat the Fundamental Theorem of Calculus actually says.

Example 2. Define a function G by

G(x) =

x

0et

2dt.

In the previous example we can actually compute x0 t

3 dt directly by finding an antiderivative oft3, but the point is that in this case this is not possible: it is not possible to write down an explicitantiderivative of et

2. (Surprising, no?) Note that here we are not saying that no one has ever found

a way to write down an explicit antiderivative, but rather that is has been proven that it is notpossible to do so.

So, this leaves us asking: how do we even know that et2has an antiderivative? The reason we

know this is true is that the Fundamental Theorem of Calculus tells us that the function G definedabove is an antiderivative of ex

2since the theorem says that

G′(x) = ex2.

Hopefully this sheds some light on why the Fundamental Theorem is even worth stating at all:there are many examples of functions for which an explicit antiderivative cannot be written down,and in such cases the only type of expression we have for such an antiderivative is the one given bythe Fundamental Theorem of Calculus itself.

Example 3. Define H by

H(x) =

x2

1t2 dt.

We want to compute H ′(x). Is the answer obtained simply by evaluating the integrand t2 as t = x2,giving H ′(x) = x4? Not quite! The point here is that the Fundamental Theorem of Calculus onlyapplies to integrals of a specific form: integrating from a constant as a lower bound to the variablex itself as the upper bound, whereas here we have an upper bound of x2. We have to be carefulwith such an expression.

To see what to do, let’s actually consider the function

F (u) =

u

1t2 dt.

I’m using u here to denote the variable instead of x so that we don’t get mixed up with H(x),which we’ll come back to in a second. The Fundamental Theorem of Calculus says that

F ′(u) = u2,

which applies since the upper bound in the integral defining F is just u and not something like u2.The relation between F and the function H we want is that H is what we get when we evaluate Fat u = x2:

H(x) = F (x2) =

x2

1t2 dt.

25

So, what we are really trying to do is compute the derivative of H(x) = F (x2). This is preciselythe type of derivative to which the chain rule applies!

Indeed, thinking of F as the “outer” function and x2 as the “inner” function, the chain rulesays that:

H ′(x) = F ′(x2) derivative of outer

(derivative of x2 derivative of inner

).

To be clear, the “derivative of outer” really means that derivative of F (u) evaluated at u = x2.Since F ′(u) = u2 by the Fundamental Theorem of Calculus, F ′(x2) = (x2)2 − x4 when we setu = x2. The derivative of x2 is 2x, so all together we get

H ′(x) = x4(2x) = 2x5

as the derivative of

H(x) =

x2

1t2 dt.

Again, the point is that in this case the upper bound is not simply x, so computing such a derivativerequires both the Fundamental Theorem of Calculus and the chain rule.

Example 4. Define

G(x) =

sinx

−1cos(t2) dt.

We want to compute G′(x), and again the answer won’t be simply what you get when you evaluatecos(t2) at t = sinx; we have to take into consideration the fact that the upper bound on the integralis sinx and not just x. Note that cos(t2) is also a function for which it is not possible to writedown an explicit antiderivative, so that in order to work with this integral we absolutely need theFundamental Theorem of Calculus Part I since the Evaluation Theorem does not apply.

Consider the function

F (u) =

u

−1cos(t2) dt.

The Fundamental Theorem givesF ′(u) = cos(u2).

The relation between F and G is that G is what you get when you plug in u = sinx into F :

G(x) = F (sinx).

The chain rule givesG′(x) = F ′(sinx)(derivative of sinx).

The first factor is F ′(u) evaluated at u = sinx, and the second is cosx, so overall we get

G′(x) = cos(sin2 x) cosx

as the derivative of G.

Example 5. Finally, we find the derivative of

F (x) =

x2

2xet dt.

26

There are a few things to note here. First, it could be that depending on x, the lower bound 2x isactually larger than the upper bound x2. (For instance, this happens when x = 1.) So, the boundson this integral seem to be in the “wrong” order. But this is simple to deal with: the notation

b

af(x) dx when b < a means by definition −

a

bf(x) dx.

In other words, to compute an integral when the bounds are in the “wrong” order we simply switchthe bounds around and change the sign of the integral. So, the expression for F (x) makes senseeven when 2x > x2.

More importantly, this integral is not the type to which the Fundamental Theorem of Calculusapplies, not only because of the x2 upper bound but also because the lower bound is not constant.(Look back to all examples we’ve said so far and note that the lower bounds were all constant,which is necessary in order to be able to apply the Fundamental Theorem.) So, the idea is that wemust rewrite the expression for F in way which involves integrals with constant lower bounds. Butthis is also simple to do: we simply introduce some intermediate constant, say 0, and break up theinterval of integration: x2

2xet dt =

0

2xet dt+

x2

0et dt.

Now, you may argue that 0 might not be between 2x and x2, so that the interval [2x, x2] is not madeup of the interval [2x, 0] together with the interval [0, x2]. This is all true, but is not a problem: ina situation such as one where

0 < 2x < x2,

we would have x2

0=

2x

0+

x2

2x,

and rearranging gives

x2

0− 2x

0=

x2

2x, which is the same as

x2

0+

0

2x=

x2

2x.

Thus, even when 0 < 2x < x2, it would still be true that

x2

2xet dt =

0

2xet dt+

x2

0et dt.

So, we never have to worry about whether or not the “intermediate” term we’re using with whichto split up an integral is actually “between” our given bounds: splitting up still works. (This is

essentially the reason why we set ba to be −

ab when a > b.)

Thus the function F can be written as

F (x) =

0

2xet dt+

x2

0et dt = −

2x

0et dt+

x2

0et dt,

and now we have the type of expressions to which the Fundamental Theorem of Calculus applies.Together with the chain rule, we thus get:

F ′(x) = −e2x(derivative of 2x) + ex2(derivative of x2)

= −2e2x + 2xex2 .

27

Lecture 5: Computing Integrals via Substitution

Warm-Up. We compute the derivative (with respect to x) of the function F defined by

F (x) =

sinx

x2tet dt.

(This is different than the specific examples we looked at in class, but it still gets the same ideaacross.) We would like to use the Fundamental Theorem of Calculus, but the given function is notin the form to which that theorem applies just yet. Our first goal is rewrite the expression definingF so that it only involves terms which look like

something depending on x

constant

since these are the types of things to which the Fundamental Theorem is applicable.First we can introduce an intermediate constant, say 2:

sinx

x2=

2

x2+

sinx

2.

(Note that here I’m focusing on the interval being integrated over, so for now I’m ignoring thefunction being integrated in the notation.) There was nothing special about 2, and we could haveinstead used sinx

x2=

10

x2+

sinx

10.

The point is that this intermediate constant isn’t going to matter anyway when we actually takethe derivative. Also note that it is not necessarily true that 2 sits between the values of x2 andsinx:

x2 < 2 < sinx,

and it could be that (depending on x) we instead have something like:

2 < sinx < x2.

We saw last time that this wasn’t an issue: in this case we could break things up as

x2

2=

sinx

2+

x2

sinx,

and then rearrange things to get sinx

x2=

2

x2+

sinx

2

as we wanted. Finally, switching the bounds changes the sign, so

sinx

x2= −

x2

2+

sinx

2,

so our defining expression for F can instead be written as

F (x) = − x2

2tet dt+

sinx

2tet dt.

28

Now we can differentiate each of these portions, using the fact that

g(x)

constantintegrand

′= integrand(g(x)) times g′(x);

that is, the derivative is obtained by evaluating the function being integrated at the upper boundg(x) and multiplying by the derivative of g(x) itself. Recall from the examples we saw last timethat this comes from the chain rule, which is needed since the upper bound is not simply x butrather some function depending on x. Ask if this is still unclear! Thus, applying this fact toeach piece of

F (x) = − x2

2tet dt+

sinx

2tet dt

givesF ′(x) = −(x2ex2)(2x) + [(sinx)esinx](cosx) = −2x3ex2 + esinx sinx cosx.

Why the Fundamental Theorem is fundamental. We now elaborate on something mentionedin passing last time. To recall, the Evaluation Theorem says that to compute an integral we shouldfirst find an antiderivative. We have said nothing about why this works until now, but the point isthat it works because of the Fundamental Theorem of Calculus! Indeed, the Fundamental Theoremprecisely says the function obtained by integrating f is a function whose derivative is f itself, sothat this integrated function is an antiderivative of f . Hence, in order to compute an integral, weshould find this antiderivative via whatever methods we have available. This is why the EvaluationTheorem is “Part II” of the Fundamental Theorem of Calculus: Part II is a consequence of Part I,which is really the key idea.

Of course, we won’t give a proof of the Fundamental Theorem of Calculus (Part I) in this course,but nonetheless here is an example which I hope illustrates why this theorem is “plausible”; thatis, why is it that differentiating

F (x) =

x

af(t) dt

should just give f(x)? Suppose f is the function whose graph is drawn below:

Recall that the function F defined above measures the net area between the graph of f and the x-axis from a up to some varying x. Now, for the function f drawn above, the Fundamental Theoremwould say that

F ′(3) = f(3) is positive.

Thinking back to what we know about derivatives, this should conceivably mean that F is increasingat x = 3, and indeed I claim this makes sense based on the graph of f ! The value of F (3) measuresthe net area up until 3, and the point is that if we then go a bit beyond x = 3 to the right, we willbe adding a positive area to this net area:

29

Thus, the net area up to a point a bit to the right of 3 will be larger than the net area up to 3,so that F evaluated at this new point will give a value larger than F (3). Hence, F does increasethrough x = 3.

Similarly, the Fundamental Theorem would say that

F ′(5) = f(5) = 0.

From what we know about derivatives, this suggests that F might have either a maximum orminimum at x = 5, and again I claim that we can tell it has a maximum based on the pictureabove. Right up to x = 5 the value of F (x) increases since up to this point we are adding positiveareas; however, once we move beyond 5 we will be adding negative areas, which will cause the valueof F to decrease:

Thus, F increases up to x = 5, then begins to decrease after, which is precisely what should happenwhen F has a maximum at x = 5

The point is that the fact that F ′(x) = f(x) is something we can make sense of based on howthe net area function F is changing; F is increasing when f is positive, F is decreasing when f isnegative, and F has a maximum or minimum when f is zero, which all together says that f doesseem to behave in a manner similar to how the derivative of F should behave.

Substitution. We now return to the problem of finding antiderivatives, which we’ll spend a goodweek or two on. The first method we’ll consider is what’s called substitution, or sometimes u-substitution based on the fact that we normally use u to denote the substitution we are making.

30

The goal of substitution is to transform a given integral into a new one in terms of some newvariable, with the hope of having this new integral be simpler to compute.

Consider for instance the integral xex

2dx.

You can check by hand that 12ex2 has derivative equal to xex

2, so 12e

x2 +C is the general antideriva-

tive of xex2:

xex2dx =

1

2ex

2+ C.

The point is that we want to have a way to figure this out without having to resort to any guesswork;namely, how exactly can we find that 12e

x2 is the correct thing to consider when starting with xex2?

Here is the process. We introduce a new variable u by setting

u = x2.

Note that with this the ex2portion of our integral becomes simply eu. Our goal is now to rewrite

the given integral in terms of u instead. Recall from what we know about the derivatives that thedifferential of u is given by:

du = 2x dx,

which is obtained by differentiating both sides of u = x2. After dividing both sides by 2 we get

1

2du = x dx.

The point is that the x dx on the right makes up the rest of the integral we are considering: theex

2term is eu, and the remaining x dx becomes 12 du. Thus we can rewrite our given integral as

xex

2dx =

1

2eu du.

This integral on the right is now straightforward to compute, where the key point is that we areintegrating with respect to the same variable u which our function is written in terms of; in otherwords, the integral

eu du is the same as that of

ex dx only that we denote the variable by u

instead of x. We have eu du = eu + C,

so all together we get xex

2dx =

1

2eu du =

1

2eu + C.

As a final step we can express the result back in terms of x by substituting back in for u: we hadoriginally set u = x2, so

xex2dx =

1

2eu + C =

1

2ex

2+ C,

agreeing with the same we gave at the start without any motivation.

Why substitution works. As a general mantra, whenever we have some differentiation technique,integrating it will give rise to some analogous integration technique. In this setting, substitution is

31

meant to be the analog of the chain rule. The key thing to is to recognize (if possible) the integralwe are wanting to compute as being of the form

f ′(g(x))g′(x) dx,

where a certain portion g′(x) looks to be the derivative of some other portion g(x). (Note in thexex

2example above, the x in front is similar to the derivative of the x2 portion.) This type of

expression shows up in the chain rule:

f(g(x))′ = f ′(g(x))g′(x),

which then gives f ′(g(x))g′(x) dx =

f(g(x))′ dx = f(g(x)) + C

after integrating.To phrase this in terms of u, set

u = g(x).

Then du = g′(x) dx, so the expression

f ′(g(x))g′(x) dx is the same as

f ′(u) du,

which works out to be f(u)+C = f(g(x))+C. The point is that after writing our original integralin terms of u, we can then simply integrate what we’re left with respect to u (forgetting at thispoint about what u actually stood for), and then plug back in for u in the end. (Later after we talka bit more about what integrals “really” mean, we’ll come back and talk about what du = g′(x) dxactually means.) Introducing u is meant to help us simplify our expression by isolating whichpart is the actually thing we have to integrate and which part comes from taking the derivative ofsomething else.

Another example. We compute 2x3 cos(5x4) dx.

The first is to recognize part of this as coming from the derivative of another part; in this case, x3

comes from the derivative of 5x4, so we set

u = 5x4.

(Note that setting u = x4 would also work, but in general to save on the amount of work we want uto take up as much as possible of our original expression; u = 5x4 leads to cos(u) where as u = x4

leads to cos(5u), so u = 5x4 takes up more of the cos(5x4) expression.) We get

du = 20x3 dx.

The cos(5x4) part becomes cosu, so we are left with the 2x3 dx part. Dividing both sides of

du = 20x3 dx

by 10 gives1

10du = 2x3 dx,

32

which is precisely what we want. Thus

2x3 cos(5x4) dx =

1

10cosu du.

This final integral (with respect to u) is 110 sinu+ C, so after plugging back in for u we get

2x3 cos(5x4) dx =

1

10sin(5x4) + C.

You can (and should!) verify that taking the derivative of 110 sin(5x4) indeed gives 2x3 cos(5x4).

Yet another example. We compute

e2

e

1

x lnxdx.

Forget the bounds for now. To use substitution we again recognize that part of our integrand comesfrom the derivative of another part; in this case, 1x comes from the derivative of the lnx in thedenominator. Thus we set

u = lnx, in which case du =1

xdx.

Thinking of our integrand as 1x1

lnx dx, we thus have1u du. Hence:

1

x lnxdx =

1

udu = ln |u|+ C = ln | lnx|+ C.

Again, you should verify that the derivative of ln | lnx| is 1x lnx . Thus now taking the bounds intoaccount, we get:

e2

e

1

x lnxdx = (ln | lnx|)

e2

e= ln(ln(e2))− ln(ln e) = ln 2− ln 1 = ln 2.

Note that above we first worked out the antiderivative, expressed everything back in terms ofx, and then used the bounds. Alternatively, once we get to

1

x lnxdx =

1

udu

we can work out new bounds on the u-integral in terms of what happens to u as x varies. In ourcase, the substitution we used was u = lnx: when x = e in the original integral, u will have thevalue u = ln e = 1, and when x = e2 in the original integral, u will have the value u = ln(e2) = 2.In other words, as x ranges from e to e2, u = lnx will in turn range from 1 to 2. This gives thebounds we want on the u-integral:

e2

e

1

x lnxdx =

2

1

1

udu.

Computing the integral on the right gives

2

1

1

udu = ln |u|

2

1= ln 2− ln 1 = ln 2,

33

which agrees with what we had before. This technique will always work: we can either computedefinite integral by expressing everything in terms of x and using the original bounds of integration,or we can keep everything in terms of u and work out the new bounds on u by seeing what happensto u when x is evaluated at the original bounds.

Final example. Finally, we compute

x√2 + x dx.

The new thing here is that there is no part of this which looks like the derivative of another part,since the x in front does not come from the derivative of the 2 + x term under the square root.Nonetheless, let us see what happens if we use the substitution

u = 2 + x

anyway. Then du = dx, so the√2 + x and dx portions of the original integral are taken care of:

they becomes√u and du respectively. But what about the remaining x in front? The point is that

we do know how to express this x in terms of u: since u = 2 + x, x = u− 2. Thus we rewrite ourentire integral, originally in terms of x, as:

x√2 + x dx =

(u− 2)

√u du.

Did this get us anywhere? Yes! Now using√u = u1/2, we can write the resulting integral as

(u− 2)u1/2 du =

(u3/2 − 2u1/2) du,

which is now possible to integrate directly. Note the key difference between this expression and theoriginal one: in the original we had

x(2 + x)1/2,

and there is no simple way to distribute the x through inside the parenthesis, whereas now we have

(u− 2)u1/2,

where distributing is possible. We end up with:

x√2 + x dx =

(u3/2 − 2u1/2) du = 2

5u5/2 − 4

3u3/2 + C =

2

5(2 + x)5/2 − 4

3(2 + x)3/2 + C.

As a check, the derivative of the expression on the right is

2

5

5

2(2 + x)3/2 − 4

3

3

2(2 + x)1/2 = (2 + x)(2 + x)1/2 − 2(2 + x)1/2 = (2 + x− 2)(2 + x)1/2,

which is x√2 + x as expected. The moral is that, really, we can make any kind of substitution we

like, as long as we can write our integral completely in terms of our new variable.

34

Lecture 6: Integration by Parts

Warm-Up 1. We compute x cos(x2)3 + sin(x2)

dx.

Note that the entire numerator comes from the derivative of sin(x2). Thus we make the substitution

u = 3 + sin(x2).

(Recall: u should take up as much of the given expression as possible, so u = 3 + sin(x2) is betterthan u = sin(x2).) Then

du = 2x cos(x2) dx,

so1

2du = x cos(x2) dx.

Thus our original integral becomes

x cos(x2)3 + sin(x2)

dx =

1

2

1√udu.

Writing 1√uas u−1/2, we have

1

2

1√udu =

1

2

u−1/2 du = u1/2 + C.

Thus x cos(x2)3 + sin(x2)

dx =√u+ C =

3 + sin(x2) + C.

As a check, taking the derivative of the function we end up with gives

1

2(3 + sin(x2))−1/2(2x cos(x2)) =

x cos(x2)3 + sin(x2)

as expected.

Warm-Up 2. We compute x3

x2 + 1 dx.

This is one of those situations where the x3 in front doesn’t come directly from the derivative ofthe term under the square root, but where we set u equal to x2 + 1 anyway and in the end writeeverything in terms of u. We have

u = x2 + 1, so du = 2x dx, and1

2du = x dx.

We need to come up with x3 dx, whereas so far we only have x dx. Thus we must multiply throughby x2, and the point is that we do know what x2 equals in terms of u: since u = x2+1, x2 = u− 1.Thus

x3 dx = x2(x dx) = (u− 1)12du.

35

Thus our original integral becomes

x3

x2 + 1 dx =

1

2(u− 1)

√u du.

Computing gives:1

2

(u3/2 − u1/2) du = 1

2

2

5u5/2 − 2

3u3/2

+ C,

so after plugging back in u = x2 + 1 we get:

x3

x2 + 1 dx =

1

5(x2 + 1)5/2 − 1

3(x2 + 1)3/2 + C.

You can (and should!) verify that taking the derivative of the expression we ended up with doesindeed give x3

√x2 + 1.

Warm-Up 3. Suppose f is a continuous function which satisfies 90 f(x) dx = 4. We compute

3

0xf(x2) dx.

The point here is that we don’t have a specific f in mind, but that the same method we’ve beenusing is applicable nonetheless. To see what to do, imagine that we did have a specific f in mind,say f(x) = sinx. Then f(x2) becomes sin(x2), so we would be looking at the integral

3

0x sin(x2) dx.

If this were the case, we can see that we should in fact make the substitution u = x2. The pointis that this is still true even when we have a unspecific f ; the x in front of xf(x2) comes from thederivative of x2, so u = x2 is the correct substitution to make.

With u = x2, we get du = 2x dx, so

1

2du = x dx.

Then our integral becomes 3

0xf(x2) dx =

1

2

9

0f(u) du,

where the new bounds come from evaluating u = x2 as the old bounds x = 0 and x = 3. Lo andbehold this resulting integral is precisely the one we are told has the value 4 at the beginning! Keepin mind that 9

0f(x) dx and

9

0f(u) du

denote the same thing, just using different notation for the variable of integration. Thus, in ourcase we get 3

0xf(x2) dx =

1

2

9

0f(u) du =

1

2(4) = 2

as the desired value.

36

Integration by parts. Suppose we want to compute

xex dx.

I claim that so far we don’t have a good way of doing this. For instance, substitution may notwork since the x in front is not the derivative of the exponent x in the ex term. (Contrast this withxex

2dx, where substitution would work nicely.) Instead we could try setting u = ex, in which

case du = ex dx. Since then x = lnu, the given integral could be written as

lnu du.

But again we’re stuck now, as we don’t have a good way of computing this integral yet either.The integration technique we now look at is called integration by parts, and is meant to transform

a given integral into a simpler to compute integral. This method is the integration-analog of theproduct rule, which says:

[f(x)g(x)]′ = f(x)g′(x) + f ′(x)g(x).

Rearranging givesf(x)g′(x)− [f(x)g(x)]′ − f ′(x)g(x),

and taking antiderivatives of both sides gives:

f(x)g′(x) dx = f(x)g(x)−

f ′(x)g(x) dx.

The left hand side is meant to denote the integral we are originally given, so this says that if wecan express the given integrand as

(function)(derivative of another function),

the formula above gives us a way to rewrite what we have.To make notation simpler, it is common to introduce the variables

u = f(x) and v = g(x),

so that du = f ′(x) dx and dv = g′(x) dx. The integration by parts formula then becomes

u dv = uv −

v du.

The starting point in using this technique is to decide what plays the role of u and what plays therole of dv. Let’s look at some examples.

Example 1. We return to the integral xex dx.

Again, we want to think of xex as the product of some function u with the derivative dv of someother function v. Let’s set

u = x and dv = ex dx.

37

Note that with these choices u dv = xex dx is indeed the thing we are integrating in the givenintegral. Since u = x we get du = dx. Now, v should be a function whose differential is dv = ex dx,or in other words v should be an antiderivative of ex; taking v = ex works. To summarize, we have:

u = x v = ex

du = dx dv = ex dx.

Plugging this into the integration by parts formula

u dv = uv −

v du

gives xex dx = xex −

ex dx.

Thus, as claimed, integration by parts has rewritten our original integral in a different form. Thepoint now is that the remaining integral

ex dx is simply to compute: it is ex+C. Thus all together

we get xex dx = xex −

ex dx = xex − ex + C

as our answer. As a check, the derivative of the resulting expression is:

(xex − ex + C)′ = xex + ex − ex = xex,

where for the first term we use the product rule. Thus xex−ex+C is indeed the general antideriva-tive of xex.

Now, why did we set u = x and dv = ex dx at the beginning, instead of say

u = ex and dv = x dx.

After all, these choices still give u dv as xex dx as we want. In this case, for du and v we get:

u = ex v =1

2x2

du = exdx dv = x dx.

Putting this into the integration by parts formula gives:

xex dx =

1

2x2ex −

1

2x2ex dx.

This is certainly a true statement, but the problem is that the remaining integralx2ex dx is now

not so simple to compute, and indeed it is more complicated than the originalxex dx we started

with! The point is that we should aim to have the remainingv du integral be simpler than the one

we started with, which happens for u = x, dv = ex dx but not u = ex, dv = x dx. The underlyingreason here is that differentiating x gives some simpler, but anti-differentiating x gives somethingmore complicated.

Example 2. We compute x2 cosx dx.

38

We setu = x2 and dv = cosx dx.

Note that instead trying to use u = cosx, dv = x2 dx would lead to a more complicated secondintegral in the integration by parts formula, since integrating x2 to get v will result in an x3 term.With the choices above we get:

u = x2 v = sinx

du = 2x dx dv = cosx dx

so integration by parts gives

x2 cosx dx = x2 sinx−

2x sinx dx.

Note that the remaining integral is indeed “simpler” than the original one in that it has a lowerpower of x.

How do we now compute this remaining integral? Via another application of integration byparts! Set

p = 2x and dq = sinx dx.

Note that I don’t want to use u and dv here since this may cause confusion with the u and dv Iused previously; with p and dq the integration by parts formula looks like:

p dq = pq −

q dp.

We get:

p = 2x q = − cosxdp = 2 dx dq = sinx dx,

so

2x sinx dx = −2x cosx−

−2 cosx dx = −2x cosx+ 2

cosx dx = −2x cosx+ 2 sinx.

Putting this back into the2x sinx dx portion of our original integration by parts application gives:

x2 cosx dx = x2 sinx−

2x sinx dx

= x2 sinx− [−2x cosx+ 2 sinx] + C= x2 sinx+ 2x cosx− 2 sinx+ C.

Again, to be clear, the2x sinx dx was itself computed via an integration by parts. To check that

this is correct, we differentiate the result using some product rules:

(x2 sinx+ 2x cosx− 2 sinx+ C)′ = 2x sinx+ x2 cosx+ 2 cosx− 2x sinx− 2 cosx = x2 cosx

as we would want to be the case.

Example 3. Finally we compute lnx dx.

39

Thinking of the integrand as lnx times 1 dx, we use

u = lnx and dv = dx.

Note that trying to use something like dv = lnx dx wouldn’t work since in order to figure out vfrom this we would have to know the antiderivative of lnx, which is precisely what it is we’re tryingto find in this problem. We get:

u = lnx v = x

du =1

xdx dv = dx,

so lnx dx = x lnx−

x1

xdx = x lnx−

dx = x lnx− x+ C.

Note what happened here: the v du portion ends up simplifying to give a simple integral to computein the

v du portion of the integration by parts formula.

You can check that differentiating the resulting expression indeed gives lnx, but recall actuallythat this is what you did on the second problem on Worksheet 1 from discussion. Now we knowhow to derive this antiderivative directly using integration by parts.

Lecture 7: Trigonometric Integrals

Warm-Up 1. We compute the definite integral

2

12x3ex

2dx

using integration by parts. First off, the bounds don’t alter the methods since we can simply putbounds everywhere in the integration by parts formula:

b

au dv = uv

b

a−

b

av du.

Now, we have to decide how we can express the expression we are integrating as one function timesthe derivative of another. We use:

u = x2 and dv = 2xex2dx.

Note that this indeed gives the correct expression for u dv, and that the v term can be found byintegrating

2xex2dx

using, say, a substitution; we get v = ex2. Thus we have:

u = x2 v = ex2

du = 2x dx dv = 2xex2dx.

Integration by parts gives:

2

12x3ex

2dx = x2ex

22

1−

2

12xex

2dx.

40

This remaining integral can be computed via a substitution, or by noting that it is precisely whatwe took to be dv, so that its integral is the v we had above. We get:

2

12x3ex

2dx = x2ex

22

1−

2

12xex

2dx

= x2ex22

1− ex2

2

1

= 4e4 − e− (e4 − e) = 3e4

as the final value.Now, what else could we have tried? Going back to the original integral, say we had thought

to tryu = 2x3 and dv = ex

2dx.

The problem with this is that ex2is a function for which it is not possible to write down an explicit

antiderivative! (Note that doesn’t mean no one has yet a found a way to do it, but that it can infact be proven that it not possible to do so.) So, we would be stuck when trying to use this as dv.This is at the heart of the reason why we set dv = 2xex

2, to get something for which v is actually

possible to compute. The problem with using something like

u = ex2and dv = 2x3 dx

is that the remaining integral after using integration by parts would actually be more complicatedthan the original one we’re trying to compute! Indeed, this remaining integral would involve x5ex

2,

which is worse than the original x3ex2. The moral is that finding the correct u and dv to use isn’t

always straightforward, and is the kind of thing which becomes simpler with practice.Nonetheless, here is another approach which also works. Let’s first start with a regular substi-

tution:u = x2 so du = 2x dx.

Then2x3ex

2dx = ueu du,

so that our original integral becomes

2

12x3ex

2dx =

4

1ueu du.

To compute the resulting integral in terms of u we can now do an integration by parts, say with

p = u and dq = eu du.

After working this out you end up with the same 3e4 value we found above. Thus in this methodwe still end up doing an integration by parts anyway, but we first use a substitution to rewrite thegiven integral as a simpler integral in terms of u.

Warm-Up 2. We compute ex cosx dx.

This will illustrate one more type of technique which might show up when doing an integration byparts. First, we set u = ex, dv = cosx dx, so that:

u = ex v = sinx

41

du = ex dx dv = cosx dx.

Integration by parts gives:

ex cosx dx = ex sinx−

ex sinx dx.

For this remaining integral we can again do an integration by parts:

p = ex q = − cosxdp = ex dx dq = sinx dx.

We get:

ex cosx dx = ex sinx−

ex sinx dx

= ex sinx−−ex cosx−

−ex cosx dx

= ex sinx+ ex cosx−

ex cosx dx.

At this point it seem as if we’re stuck, since if keep doing integration by parts we’ll just keepflipping back and forth between ex sinx integrals and ex cosx integrals. However, here is the keyobservation: the remaining

ex cosx dx integral is precisely the integral on the left side we were

originally trying to compute. Think of

ex cosx dx = ex sinx+ ex cosx−

ex cosx dx

as an equation whereex cosx dx is the unknown quantity we want to find. In other words, if I

think ofex cosx dx as just some unknown variable:

=

ex cosx dx,

then the equation we have is = ex sinx+ ex cosx− .

We can solve this for by adding it to both sides to get:

2 = ex sinx+ ex cosx.

In terms of what represents this says

2

ex cosx dx = ex sinx+ ex cosx,

so dividing by 2 gives the value we’re looking for:

ex cosx dx =1

2(ex sinx+ ex cosx) + C.

(The +C was thrown on at the end as usual.) We can verify that this is the correct indefiniteintegral by taking the derivative of what we got:

1

2(ex sinx+ ex cosx)′ =

1

2(ex sinx+ ex cosx+ ex cosx− ex sinx) = ex cosx.

42

Trigonometric Integrals. We now consider integrals which involve powers of the basic trigfunctions sinx, cosx, secx, tanx. Such things are called trigonometric integrals, and the point hereis that we can use well-known trig identities to transform such integrals into more easily computableforms. The key trig identities we use are:

sin2 θ + cos2 θ = 1, sec2 θ = tan2 θ + 1, cos2 θ =1

2(1 + cos 2θ), sin2 θ =

1

2(1− cos 2θ).

The first is one we all know and love, the second comes from dividing both sides of the first bycos2 θ, and the second and third are called the half-angle formulas. Note that these last two turnexpressions involving sine or cosine squared into ones involving sine and cosine themselves with theangle doubled.

Example 1. We compute sin3 x dx.

First note that normal substitution u = sinx doesn’t work since this would give a cosx term in du,but there is no cosx term in the given integral which we would need to be able to introduce du. (Ifthis was

sin3 x cosx dx instead then a simple substitution u = sinx would indeed work nicely.)

The key here is that we can think of sin3 x (which is sinx times itself three times) as

sin3 x = sin2 x sinx,

and the sin2 x term is something we can replace with something involving cos2 x, namely 1−cos2 x.We have:

sin3 x dx =

sin2 x sinx dx =

(1− cos2 x) sinx dx.

Now the resulting integral is one we can do using a substitution, say u = cosx. With this we getdu = − sinx dx, so the given integral becomes

sin3 x dx =

(1− cos2 x) sinx dx

=

−(1− u2) du

= −u+ 13u3 + C

= − cosx+ 13cos3 x+ C.

To check that this is correct, we compute a derivative:

− cosx+ 1

3cos3 x

′= sinx+

3

3cos2 x(− sinx) = sinx(1− cos2 x) = sinx sin2 x = sin3 x,

so that − cosx+ 13 cos3 x+ C is indeed the general antiderivative of sin3 x.

The goal here and for general trig integrals is to use a trig identity to rewrite the given integralin a different form, where one portion ends up being related to the derivative of another portion,after which a normal substitution should come in handy.